In △ABC,∠A=x∘,∠B=(2x-15)∘and ∠C=(3x+21)∘. Find the value of x and the measure of each angle of the triangle.

x = 29 ∘ , ∠ A = 29 ∘ , ∠ B = 43 ∘ , ∠ C = 108 ∘

x = 29 ∘ , ∠ A = 29 ∘ , ∠ B = 43 ∘ , ∠ C = 108 ∘

x = 27 ∘ , ∠ A = 27 ∘ , ∠ B = 41 ∘ , ∠ C = 101 ∘

x = 29 ∘ , ∠ A = 29 ∘ , ∠ B = 43 ∘ , ∠ C = 108 ∘

x = 27 ∘ , ∠ A = 27 ∘ , ∠ B = 41 ∘ , ∠ C = 108 ∘

x = 30 ∘ , ∠ A = 30 ∘ , ∠ B = 41 ∘ , ∠ C = 109 ∘

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In $△ ABC, ∠ A = x^{\circ}, ∠ B = {(2 x - 15)}^{\circ}$ and $∠ C = {(3 x + 21)}^{\circ}$ . Find the value of $x$ and the measure of each angle of the triangle.

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x = 27^{\circ}, ∠ A = 27^{\circ}, ∠ B = 41^{\circ}, ∠ C = 101^{\circ}

x = 29^{\circ}, ∠ A = 29^{\circ}, ∠ B = 43^{\circ}, ∠ C = 108^{\circ}

x = 27^{\circ}, ∠ A = 27^{\circ}, ∠ B = 41^{\circ}, ∠ C = 108^{\circ}

x = 30^{\circ}, ∠ A = 30^{\circ}, ∠ B = 41^{\circ}, ∠ C = 109^{\circ}

answer is B.

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Given,

∠ A = x^{\circ}

∠ B = (2 x - 15)^{\circ}

and

∠ C = (3 x + 21)^{\circ}

.
We know, by angle sum property, the sum of angles of a triangle is

180^{\circ}

.Then,

∠ A + ∠ B + ∠ C = 180^{\circ}

⟹

x^{\circ} + (2 x - 15)^{\circ} + (3 x + 21)^{\circ} = 180^{\circ}

⟹

6 x^{\circ} + 6^{\circ} = 180^{\circ}

⟹

6 x^{\circ} = 180^{\circ} - 6^{\circ}

⟹

6 x^{\circ} = 174^{\circ}

⟹

x^{\circ} = 29^{\circ}

Therefore,

∠ A = x^{\circ} = 29^{\circ}

∠ B = (2 x - 15)^{\circ}

⟹ ∠ B = 2 \times 29^{\circ} - 15^{o}

⟹ ∠ B = 58^{\circ} - 15^{\circ}

⟹ ∠ B = 43^{\circ}

∠ C = (3 x + 21)^{\circ}

⟹ ∠ C = 3 \times 29^{\circ} + 21^{\circ}

⟹ ∠ C = 87^{\circ} + 21^{\circ}

⟹ ∠ C = 108^{\circ}

Thus, the values of

x = 29^{\circ}, ∠ A = 29^{\circ}, ∠ B = 43^{\circ}, ∠ C = 108^{\circ}

Therefore, option

2

is correct.

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